Assume that \(\left( \Phi _{(n-1)f}^{\ast }(p),k+1\right) >1\) . If \(X\cong PSU_{n}(q^{1/2})\) , with \(n\)  even, then either
\(n=4\)  and \(PSL_{3}(2^{t})\leqslant X_{x}\)  with \((t,q)=(1,3),(1,5),(2,3)\) , or \(X_{x}\) 
preserves either a point or a hyperplane of \(PG_{n-1}(q)\)  by [1]},
Theorem 4.2. If \((t,q)=(2,3)\) , then \(X_{x}\cong PSL_{3}(4)\)  by [2]} and
hence \(k^{2}=162\) , a contradiction. If \((t,q)=(1,3)\) , then \(PSL_{3}(2)\leqslant X_{x}\leqslant M\) , where \(M\)  is one of the following maximal subgroups of \(X\)  by [2]}:
\(PSL_{3}(4)\) , \(PSU_{3}(3)\)  or \(A_{7}\) . Either \(X_{x}\cong PSL_{3}(2)\)  or \(X_{x}=M\) , since \(PSL_{3}(2)\)  is maximal in \(M\) .
Actually, the unique admissible case is \(X_{x}\cong A_{7}\)  and \(k=36\) , since \(\left[ X:X_{x}\right] =k^{2}\) . However, this case is ruled out, since \(k+1\)  does not divide the order of \(P\Gamma U_{4}(3)\) , and hence that of \(G\) .
Assume that \(X_{x}\)  preserves either a point or a hyperplane of \(PG_{n-1}(q)\) . Actually, \(G_{x}\)  is the stabilizer of a non-isotropic point \(P\)  of \(PG_{n-1}(q)\)  by Proposition REF  and by [4]}, Proposition 4.1.6.(II), since \(\Phi _{ef}^{\ast }(p)\mid \left|X_{x}\right|\) . Hence, the actions of \(G\)  on the point set of \(\mathcal {D}\) 
and on the set of non-singular points of \(PG_{n-1}(q)\)  are equivalent. Thus, \(k^{2}=q^{\left( n-1\right) /2}\frac{q^{n/2}-1}{q^{1/2}+1}\)  and hence \(f\equiv 0 \pmod {4}\)  as \(q^{\left( n-1\right) /2}\)  must be a
square, being \(n\)  is even. Then
\(k^{2}=s^{2\left( n-1\right) }\frac{\left( s^{n}+1\right) \left(s^{n}-1\right) }{s^{2}+1}\text{,} \) 
where \(s=q^{1/4}\) . Then \(s\)  is odd by [5]}, A3.1. Moreover, if we consider a
further non-singular point \(y\)  of \(PG_{n-1}(q)\)  such that \(y\in x^{\perp }\) , we see that \(\left|y^{G_{x}}\right|\)  divides \((s^{2}+1)(s^{2(n-1)}+1)\) . Then \(k+1\mid \)  \((s^{2}+1)(s^{2(n-1)}+1)\)  by Lemma REF (3). Actually, \(k+1\mid \)  \(\frac{(s^{2}+1)(s^{2(n-1)}+1)}{4}\)  as \(s\) 
is odd, which, compared to (REF ), leads to a contradiction as \(n \geqslant 4\) . Thus, \(X\ncong PSU_{n}(q^{1/2})\)  for \(\left( \Phi _{(n-1)f}^{\ast }(p),k+1\right) >1\) .
Assume that \(X\cong P\Omega _{n}(q)\) , with \(nq\)  odd. Assume also that \(\left|X_{x}\right|_{p}<\left|\mathrm {Out(}X\mathrm {)}\right|\) . Then
\(\left|X_{x}\right|_{p}<2f\) , as \(\left|\mathrm {Out(}X\mathrm {)}\right|=2f\) . Since \(X_{x}<X<PSL_{n}(q)\)  and \(\Phi _{(n-1)f}^{\ast }(p)\)  divides \(\left|X_{x}\right|\) , it follows that \(X_{x}^{^}\)  is one of the groups listed in Theorem 3.1 of [1]}.
Assume that \(X_{x}^{^}\)  is a nearly simple subgroup of \(SL_{n}(q)\) . Suppose that \(\mathrm {Soc(}X_{x}\mathrm {)}\)  is alternating. Then \(X_{x}\) 
arises from the action of \(A_{\ell }\)  on its fully deleted permutation
module, since \(nq\)  is odd. Then \(q\)  is prime, \(n=\ell -1\)  or \(\ell -2\) 
according as \(q\)  does not divide or does divide \(n\)  respectively. In both cases, we have \(\Phi _{n-1}^{\ast }(q)=n\) , with \(n=4,6\) , whereas \(n\)  is odd.
Assume that \(\mathrm {Soc(}X_{x}\mathrm {)}\)  is a sporadic group. Then \(\mathrm {Soc(}X_{x}\mathrm {)}\cong M_{11}\trianglelefteq X_{x}\) , \(n=5\)  and \(q=3\) . So \(\left|X_{x}\right|_{3}=3^{2}\) , whereas \(\left|\mathrm {Out(}X\mathrm {)}\right|=2\) .
Assume that \(\mathrm {Soc(}X_{x}\mathrm {)}\)  is a simple Lie type group in
cross characteristic. \(\left|X_{x}\right|_{p}<\left|\mathrm {Out(}X\mathrm {)}\right|\) 
implies \(n=7\) , \(q=5\)  and \(\mathrm {Soc(}X_{x}\mathrm {)}\)  isomorphic to one of
the groups \(PSL_{2}(7)\) , \(PSL_{2}(8)\)  or \(PSU_{3}(3)\) . However, these cases
are ruled out. Indeed, 31 must divide \(\left|X_{x}\right|\) , since \(\left|P\Omega _{7}(5)\right|\)  is divisible by 31, and not by \(31^{2}\) , and since \(\left[X:X_{x}\right] =k^{2}\) .
Assume that \(\mathrm {Soc(}X_{x}\mathrm {)}\)  is a simple group of Lie type in
characteristic \(p\) . Then \(n=7\) , \(q\)  is odd, with \(q\)  dividing \(\left|X_{x}\right|\)  by Theorem 3.1 of [1]}. So \(q<\left|\mathrm {Out(}X\mathrm {)}\right|=2f\) , which has no solutions for \(q\)  odd.
Clearly \(X_{x}^{^}\)  is not a Symplectic Type Example. Also, \(X_{x}^{^}\)  is not a Classical Example since \(X_{x}<X\cong P\Omega _{n}(q)\) . If \(X_{x}^{^}\)  is an Imprimitive Example, then \(q=p\)  and
\(X_{x}^{^}\leqslant GL_{1}(p)\wr S_{n}\) , where either \(q=3\)  and \(n=5,7\) ,
or \(q=5\)  and \(n=7\)  by Theorem 3.1 of [1]}, being \(n\)  odd. The case \(q=3\)  and \(n=5\)  cannot occur, since \(P\Omega _{5}(3)\cong PSU_{4}(2)\)  has
been ruled out above. The groups \(P\Omega _{7}(3)\)  and \(P\Omega _{7}(5)\)  are
ruled out, as they do not have a transitive permutation representation of
square degree by [9]}, Tables 8.39 and 8.40. Therefore, \(X_{x}\)  fixes
either a point or a hyperplane of \(PG_{n-1}(q)\) . So does \(G_{x}\) , as \(X_{x}\trianglelefteq G_{x}\) . Then \(G_{x}\)  should be one of the groups listed in [10]}, Lemma
3.13, since \(\left|X_{x}\right|_{p}<\left|\mathrm {Out(}X\mathrm {)}\right|\) , whereas it is not, and we obtain a contradiction. Thus, \(\left|X_{x}\right|_{p}\geqslant \left|\mathrm {Out(}X\mathrm {)}\right|\) 
and hence \(X_{x}\)  is a large subgroup of \(X\)  by Lemma REF (2).
